Maximal Rectangle

最新推荐文章于 2021-10-17 09:55:57 发布

原创最新推荐文章于 2021-10-17 09:55:57 发布 · 246 阅读

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本文介绍了一种求解二维矩阵中最大全1矩形面积的算法。通过预处理计算每行的高度数组，并利用单调栈来寻找以每行作为底边的最大矩形面积，最终得出整个矩阵中的最大矩形面积。

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public class Solution {
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
            return 0;
        }
        int r = matrix.length, c = matrix[0].length;
        int[][] heights = new int[r][c];
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                if (i == 0) {
                    heights[i][j] = matrix[i][j] - '0';
                } else {
                    if (matrix[i][j] != '0') {
                        heights[i][j] = 1 + heights[i - 1][j];
                    }   
                }
            }
        }
        int maxArea = 0;
        for (int i = 0; i < r; i++) {
            Stack<Integer> stack = new Stack<>();
            for (int j = 0; j < c; j++) {
                if (stack.isEmpty()) {
                    stack.push(j);
                } else {
                    //int h = heights[i][stack.peek()];
                    while (!stack.isEmpty() && heights[i][j] < heights[i][stack.peek()]) {
                        //int maxH = stack.pop();
                        int maxH = heights[i][stack.pop()];
                        if (!stack.isEmpty()) {
                            maxArea = Math.max(maxArea, (j - 1 - stack.peek())*maxH);
                        } else {
                            maxArea = Math.max(maxArea, (j)*maxH);
                        }
                    }
                    stack.push(j);
                }
            }
            while (!stack.isEmpty()) {
                int h = heights[i][stack.pop()];
                if (!stack.isEmpty()) {
                    maxArea = Math.max(maxArea, (c - 1 - stack.peek())*h);
                } else {
                    maxArea = Math.max(maxArea, c*h);
                }
            }
        }
        return maxArea;
    }
}